\begin{tabular}{c|} If $$ y=15-\frac{1}{x^{6}}+\frac{1}{3} x^{20} $$, then $$ y^{\prime}=\ldots \ldots \ldots \ldots $$ \ $$ \frac{6}{x^{7}}+\frac{20}{3} x^{19} $$ \ $$ \frac{6}{x^{5}}+\frac{20}{3} x^{19} $$ \ $$ \frac{6}{x^{5}}+\frac{20}{3} x^{19}+15 $$\end{tabular}

Question

UpStudy AI · Accepted Answer

We are given 
$$
y=15-\frac{1}{x^6}+\frac{1}{3}x^{20}.
$$
We will differentiate each term with respect to $$ x $$.

1. The derivative of the constant $$ 15 $$ is 
   $$
   0.
   $$

2. Rewrite the term $$-\frac{1}{x^6}$$ as 
   $$
   -x^{-6}.
   $$
   Its derivative using the power rule is
   $$
   \frac{d}{dx}\left(-x^{-6}\right) = -(-6)x^{-7} = 6x^{-7} = \frac{6}{x^7}.
   $$

3. The term $$\frac{1}{3}x^{20}$$ differentiates as follows:
   $$
   \frac{d}{dx}\left(\frac{1}{3}x^{20}\right)=\frac{1}{3}\cdot 20x^{19} = \frac{20}{3}x^{19}.
   $$

Thus, combining these results, the derivative is
$$
y' = \frac{6}{x^7}+\frac{20}{3}x^{19}.
$$
The derivative $$ y' $$ is $$ \frac{6}{x^7} + \frac{20}{3}x^{19} $$.

\begin{tabular}{c|} If \( y=15-\frac{1}{x^{6}}+\frac{1}{3} x^{20} \), then \( y^{\prime}=\ldots \ldots \ldots \ldots \) \\ \( \frac{6}{x^{7}}+\frac{20}{3} x^{19} \) \\ \( \frac{6}{x^{5}}+\frac{20}{3} x^{19} \) \\ \( \frac{6}{x^{5}}+\frac{20}{3} x^{19}+15 \)\end{tabular}

Upstudy AI Solution

Answer

Solution

Extra Insights

Related Questions

Latest Calculus Questions